chap06

# chap06 - Chapter 6 Solution 1 i=C dv = 5 2e-3t 6 e-3 t =...

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Chapter 6, Solution 1. ( = + = = t 3 t 3 e 6 e 2 5 dt dv C i ) 10(1 - 3t)e -3t A p = vi = 10(1-3t)e -3t 2t e -3t = 20t(1 - 3t)e -6t W Chapter 6, Solution 2. 2 2 1 1 ) 120 )( 40 ( 2 1 Cv 2 1 w = = w 2 = 2 2 1 ) 80 )( 40 ( 2 1 2 = Cv 1 ( ) = = = 2 2 2 1 80 120 20 w w w 160 kW Chapter 6, Solution 3. i = C = = 5 160 280 10 x 40 dt dv 3 480 mA Chapter 6, Solution 4. ) 0 ( v idt C 1 v t o + = + 1 tdt 4 sin 6 2 1 = = 1 - 0.75 cos 4t Chapter 6, Solution 5. v = + t o ) 0 ( v idt C 1 For 0 < t < 1, i = 4t, = t o 6 t 4 10 x 20 1 v dt + 0 = 100t 2 kV v(1) = 100 kV

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For 1 < t < 2, i = 8 - 4t, + = t 1 6 ) 1 ( v dt ) t 4 8 ( 10 x 20 1 v = 100 (4t - t 2 - 3) + 100 kV Thus v (t) = < < < < 2 t 1 , kV ) 2 t t 4 ( 100 1 t 0 , kV t 100 2 2 Chapter 6, Solution 6. 6 10 x 30 dt dv C i = = x slope of the waveform. For example, for 0 < t < 2, 3 10 x 2 10 dt dv = i = mA 150 10 x 2 10 x 10 x 30 dt dv 3 6 = = C Thus the current i is sketched below. t (msec) 150 12 10 2 8 6 4 -150 i(t) (mA) Chapter 6, Solution 7. + = + = t o 3 3 o 10 dt 10 tx 4 10 x 50 1 ) t ( v idt C 1 v = = + 10 50 t 2 2 0.04k 2 + 10 V
Chapter 6, Solution 8. (a) t t BCe ACe dt dv C 600 100 600 100 = = i (1) B A BC AC i 6 5 600 100 2 ) 0 ( = → = = (2) B A v v + = → = + 50 ) 0 ( ) 0 ( (3) Solving (2) and (3) leads to A=61, B=-11 (b) J 5 2500 10 4 2 1 ) 0 ( 2 1 3 2 = = = x x x Cv Energy (c ) From (1), A 4 . 26 4 . 24 10 4 11 600 10 4 61 100 600 100 600 3 100 3 t t t t e e e x x x e x x x i = = Chapter 6, Solution 9. v(t) = ( ) ( ) + = + t o t t V e t 12 0 dt e 1 6 2 1 1 v(2) = 12(2 + e -2 ) = 25.62 V p = iv = 12 (t + e -t ) 6 (1-e -t ) = 72(t-e -2t ) p(2) = 72(2-e -4 ) = 142.68 W Chapter 6, Solution 10 dt dv x dt dv C i 3 10 2 = = < < < < < < = s 4 t 3 16t, - 64 s 3 t 1 16, s 1 0 , 16 µ µ µ t t v

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< < < < < < = s 4 t 3 , 16x10 - s 3 t 1 0, s 1 0 , 10 16 6 6 µ µ µ t x dt dv < < < < < < = s 4 t 3 kA, 32 - s 3 t 1 0, s 1 0 , kA 32 ) ( µ µ µ t t i Chapter 6, Solution 11. v = + t o ) 0 ( v idt C 1 For 0 < t < 1, = = t o 3 6 t 10 dt 10 x 40 10 x 4 1 v kV v(1) = 10 kV For 1 < t < 2, kV 10 ) 1 ( v vdt C 1 v t 1 = + = For 2 < t < 3, + = t 2 3 6 ) 2 ( v dt ) 10 x 40 ( 10 x 4 1 v = -10t + 30kV Thus v(t) = < < + < < < < 3 t 2 , kV 30 t 10 2 t 1 , kV 10 1 t 0 , kV t 10 Chapter 6, Solution 12. π π = = 4 sin )( 4 ( 60 x 10 x 3 dt dv C i 3 t) = - 0.7e π sin 4 π t A P = vi = 60(-0.72) π cos 4 π t sin 4 π t = -21.6 π sin 8 π t W W = t dt π π = t o 8 1 o 8 sin 6 . 21 pdt = π π π 8 8 6 . cos 21 8 / 1 o = -5.4J
Chapter 6, Solution 13. Under dc conditions, the circuit becomes that shown below: i 2 50 20 + 60V + v 1 i 1 30 10 + v 2 i 2 = 0, i 1 = 60/(30+10+20) = 1A v 1 = 30i 2 = 30V, v 2 = 60-20i 1 = 40V Thus, v 1 = 30V, v 2 = 40V Chapter 6, Solution 14. (a) C eq = 4C = 120 mF (b) 30 4 C 4 C 1 eq = = C eq = 7.5 mF Chapter 6, Solution 15. In parallel, as in Fig. (a), v 1 = v 2 = 100 C 2 + v 2 C 1 + 100V + v 2 C 2 + v 1 C 1 + v 1 + 100V (b) (a)

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w 20 = = = 2 6 2 100 x 10 x 20 x 2 1 Cv 2 1 0.1J w 30 = = 2 6 100 x 10 x 30 x 2 1 0.15J (b) When they are connected in series as in Fig. (b): , 60 100 x 50 30 V C C C v 2 1 2 1 = = + = v 2 = 40 w 20 = = 2 6 60 x 10 x 30 x 2 1 36 mJ w 30 = = 2 6 40 10 x 30 2 x x 1 24 mJ Chapter 6, Solution 16 F 20 30 80 80 14 µ = → = + + = C C Cx C eq Chapter 6, Solution 17.
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